求解几道英文化学题，在线等，急！

Question

1).Write a balanced equation for the reaction of calcium phosphide, Ca3P2, with
water, H2O, forming calcium hydroxide, Ca(OH)2 and phosphine, PH3. Calculate the
heat of reaction from the standard heats of formation listed below. Then,
calculate the standard enthalpy change when 39.4 grams of calcium phosphide is
consumed in this reaction. standard heat of formation(Ca3P2) -504 kJ/mol
standard heat of formation(H2O) -285 kJ/mol standard heat of formation(Ca(OH)2)
-986 kJ/mol standard heat of formation(PH3) 5 kJ/mol
2).If 2.765×10-2 mol of NH3 gas has a volume of 715 mL at
42°C, what would be the volume of 2.765×10-2 mol of N2 gas
under identical temperature and pressure conditions?
3). A sample of
N2H6CO2 weighing 14.5 grams is heated to
80oC in a 1.0 L container. All of the salt sublimes into
NH3(g) and CO2(g) according to the following equilibrium
process.

N2H6CO2 (s) ↔ CO2 (g) + 2NH3 (g)

Determine the partial
pressure of the NH3(g) in the system in atmospheres.

Answer 1

1）请为磷化钙，ca3p2反应平衡方程，用水，水，形成氢氧化钙，Ca（OH）2和膦，PH3。计算
从形成下面列出的标准生成热反应热。然后，标准焓计算时，39.4克磷化钙在该反应消耗。标准生成热（ca3p2）- 504 kJ / mol标准生成热（H2O）- 285 kJ / mol的标准生成热（Ca（OH）2）
- 986 kJ / mol的标准生成热（PH3）5 kJ / mol
2）。如果2.765×10-2 mol NH3气体的体积715毫升42°C，什么是2.765×10-2 mol N2气体的体积
相同的温度和压力条件下？
3。一个样本n2h6co2重14.5克，加热在一个1升的容器的冰箱。所有的盐升华成NH3（G）和CO2（g）根据以下的平衡过程。
n2h6co2（S）↔CO2（g）+ 2nh3（G）确定局部的NH3的压力（G）在大气中的系统。

Ca3P2+6H2O=3Ca(OH)2+2PH3↑

Answer 2

1) Ca3P2 + 6 H2O = 3Ca(OH)2 + 2PH3 + 734 kJ/mol
158.9 kJ
2) 715mL
3) 如楼上所述

Answer 3

2. pv=nRT
constant P and T, 体积与物质的量成正比，物质的量相同，所以体积也相同，715mL
3. n=m/M(相对分子质量)
先求出N2H6CO2的物质的量0.226mol
Pv=nRT
求反应后体系的总压强19.65atm
P=P总* X（摩尔分数）=13.1atm